Barber Paradox/Resolution 1

Paradox

There exists a community, one of whose members is a barber.

This barber operated under an unusual rule: his task was to shave every man in the community who did not shave himself, and only those men.

Who shaves the barber?

If he does not shave himself, then he must shave himself.

But if he shaves himself, he must not shave himself.

This is an application of Russell's Paradox.

Let $\mathbb U$ be the set of all the men of the community.

Thus $\mathbb U$ is considered to be the universe.

Let $S: \mathbb U \to \set {T, F}$ be the propositional function:

$\forall x \in \mathbb U: \map S x \iff x \text { is shaved by $x$}$

Let $b \in \mathbb U$ be the barber.

Let $B: \mathbb U \to \set {T, F}$ be the propositional function:

$\forall x \in \mathbb U: \map B x \iff x \text { is shaved by $b$}$

The initial premises can be coded:

$(1): \quad \forall x \in \mathbb U: \paren {\neg \map S x} \iff \map B x$

$(2): \quad \map B b \iff \map S b$

Hence:

$\map S b \iff \map B b \iff \paren {\neg \map S b}$

$\map S b \iff \paren {\neg \map S b}$

So from either case there derives a contradiction.

Thus the initial premises are contradictory and cannot both hold.

$\blacksquare$

Let $b$ be defined so that $b \notin \mathbb U$.

That is, suppose $b$ is not one of the men of the community.

This could be the case by, for example:

$(1): \quad$ The barber is a woman

$(2): \quad$ The barber is a boy too young to shave.

Then as $b \notin \mathbb U$, it is not necessarily the case that:

$\paren {\neg \map S b} \implies \map B b$

Thus $b$ is allowed not to be shaved, by himself or anyone else.

$\blacksquare$